Giải thích các bước giải:

a.Xét $\Delta MAB,\Delta MCD$ có:

$MA=MD$

$\widehat{AMB}=\widehat{CMD}$
$MB=MC$

$\to \Delta AMB=\Delta DMC(c.g.c)$

b.Xét $\Delta AMC,\Delta BMD$ có:

$MA=MD$

$\widehat{AMC}=\widehat{BMD}$
$MC=MB$

$\to \Delta AMC=\Delta DMB(c.g.c)$

$\to AC=BD, \widehat{MAC}=\widehat{MDB}\to AC//BD$

c.Xét $\Delta ABF,\Delta ACE$ có:

$AB=AE$

$\widehat{BAF}=90^o+\widehat{BAC}=\widehat{EAC}$

$AC=AF$

$\to\Delta ABF=\Delta AEC(c.g.c)$

$\to BF=CE, \widehat{ABF}=\widehat{AEC}$

Gọi $AB\cap CE=G, CE\cap BF=I$

$\to \widehat{AEG}=\widehat{GBI}$

$\to \widehat{GIB}=180^o-\widehat{GBI}-\widehat{BGI}=180^o-\widehat{GEA}-\widehat{AGE}=\widehat{GAE}=90^o$

$\to BF\perp CE$

d.Xét $\Delta AEF, \Delta ABD$ có:

$BD=AF(=AC)$

$\widehat{ABD}=180^o-\widehat{BAC}=\widehat{EAF}$

$AB=AE$

$\to \Delta ABD=\Delta EAF(c.g.c)$

$\to \widehat{AEF}=\widehat{BAD}$

Gọi $AD\cap EF=H'$

$\to \widehat{AEH'}=\widehat{BAD}$

$\to \widehat{AEH'}+\widehat{H'AE}=\widehat{BAD}+\widehat{H'AE}=180^o-\widehat{BAE}=90^o$

$\to \Delta H'AE$ vuông tại $H'$

$\to AH'\perp EF$

$\to H,H'$ trùng nhau

$\to A, M, H$ thẳng hàng